A function with unexpectedly simple Legendre transformation

Question

Let $I(x) = \frac{1}{2\pi} \int_{-2}^2 \sqrt{4-y^2}\ln|x-y|dy$. Then $I(x)$ is a concave function and
\begin{equation} I(x)= \begin{cases} \frac{1}{4}x^2-\frac{1}{2}, &\text{if } |x|\leq2 \\ \frac{1}{4}x^2-\frac{1}{2}-\frac{1}{4}|x|\sqrt{x^2-4}+\ln \frac{|x|+\sqrt{x^2-4}}{2} &\text{if } |x|>2 \end{cases} \end{equation} Let $J(x)$ be the Legendre transform of $\frac{1}{2}x^2 - I(x)$, i.e. $J(x): = \sup_{y\in \mathbb{R}}\{xy - \frac{1}{2}y^2 + I(y)\}$, then $J(x)$ has a quite simple form \begin{equation} J(x) = \begin{cases} x^2-\frac{1}{2} &\text{if } |x|\leq1 \\ \frac{1}{2}x^2 + \ln |x| &\text{if } |x|>1 \end{cases} \end{equation} The simple form of $J(x)$ suggests some hidden deeper truth. My question is, can we find $J(x)$ directly without calculating $I(x)$ explicitly?

Answer 1

Claim. $J(x)=\tfrac12x^2+\ln|x|+c$ for $|x|>1$ with $c$ constant.

Proof: Here, an explicit form for $I(x)$ is not needed but I can't use it to prove $c=0$.

For $|x|>1$, the substitution $y=2\cos t$ followed by $(x-\cos t)(x-\cos s)=x^2-1$ yields $$I'(2x)=\frac1{2\pi}\int_{-2}^2\frac{\sqrt{4-y^2}}{|2x-y|}\,dy=x-\frac1\pi\int_0^\pi\frac{x^2-1}{|x-\cos t|}dt=x-\sqrt{x^2-1}$$ where $'=d/dx$. Thus $I'(x)+1/I'(x)=x$ for $|x|>2$.

We have $J(x)=xy_*-\tfrac12y_*^2+I(y_*)$ where $x-y_*+I'(y_*)=0$. This is equivalent to $y_*-x+\frac1{y_*-x}=y_*$ using the above identity, which implies $y_*=x+\frac1x$. Thus $J(x)=\frac12x^2-\frac1{2x^2}+I(x+\frac1x)$ so we now evaluate $I(x+\frac1x)$.

Let $K(x)=xI^*(x+\frac1x)$ where $^*=d/d(x+\frac1x)$ so the above identity simplifies to $K(x)^2-(x^2+1)K(x)+x^2=0$. By inspection, we have $K(x)=1$ or $x^2$; however, if $K(x)=x^2$ then $I'$ behaves like $x$ asymptotically, which is a contradiction. Thus $K(x)=1$, so $I^*(x+\frac1x)=\frac1x$ implies $I'(x+\frac1x)=\frac1x(1-\frac1{x^2})$ or that $I(x+\frac1x)=\frac1{2x^2}+\ln|x|+c$. This holds for $|x|>1$ so $J(x)=\tfrac12x^2+\ln|x|+c$ for $|x|>1$.